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I came across the following expression:

$$\frac{\partial^{i_1+\cdots+i_m}P(x_1,\ldots,x_m)}{\partial x_1^{i_1}\cdot\cdot\cdot \partial x_m^{i_m}}$$

for $P(x_1,\ldots,x_m)$ a polynomial in $m$ variables $x_1,\ldots,x_m$, and I must say that I find it rather confusing, as I've never encountered this notation before.

So my question is: Given a monomial of the form $x_1^{j_1}\cdots x_m^{j_m}$, what does $\dfrac{\partial^{i_1+\,\cdots\,+i_m} x_1^{j_1} \cdots x_m^{j_m} }{\partial x_1^{i_1}\cdots \partial x_m^{i_m}}$ look like?

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I'm curious: Where did you come across that expression? – draks ... Apr 21 '12 at 20:10
@sssss : Can you tell me where you learned to use $\TeX$ for posting in this forum. Lot's of people who post here do the same thing you did: they write things like {{a}^{2}} instead of a^2, and you did lots of things like that here. I changed it since I don't want to encourage newbies to think things like that are needed. – Michael Hardy Apr 21 '12 at 20:52

Each of the $x_i$ are independent and so are constants wrt each other. We can thus separate:

$$\frac{\partial^{~i_1+\cdots+i_m}}{\partial x_1^{\,i_1}\cdots \partial x_m^{\,i_m}}x_1^{j_1}\cdots x_m^{j_m} = \left(\frac{\partial^{\,i_1}}{\partial x_1^{i_1}}x^{j_1}\right)\cdots\left(\frac{\partial^{\,i_m}}{\partial x_m^{i_m}}x^{j_m}\right). \tag{$\circ$}$$

We can evaluate each piece individually using one-variable calculus and the falling factorial:

$$\frac{\partial^{\,i}}{\partial x^i} x^j=j(j-1)\cdots\big(j-(i-1)\big) x^{j-i}=(j)_i x^{j-i}.$$

How do we know the separation in $(\circ)$ is valid? Consider induction on the following two facts:

  • If $x_1,\cdots$ are independent variables, the partial derivatives $\partial/\partial x_1,\cdots$ all commute.
  • We have $$\frac{\partial}{\partial u} [f(x_1,\cdots,x_n)g(u)]=f(x_1,\cdots,x_n) \frac{\partial}{\partial u} g(u).$$
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I think that the $x$'s in your second row should be $x_1,...,x_m$, am I right? – sssss Apr 21 '12 at 18:22
@sssss: There's no need to bother with indices as it's a generic formula for one-variable functions. Any such formula will hold no matter what the one variable in particular is, so we can replace $x,i,j$ with $x_k,i_k,j_k$ for $1\le k\le m$ if we want and it still works. – anon Apr 21 '12 at 18:24
$^\dagger$Disclaimer: pathological counterexamples to full commutativity exist outside the realm of nice behavior and everyday regularity. – anon Apr 21 '12 at 19:19


This notation using multindices ($\alpha=(\alpha_1,\ldots,\alpha_m)\in\mathbb{N}_0$ is called a multindex) is due to Laurent Schwartz. Its abbreviated form is: $$\frac{\partial^{\alpha_1}}{\partial x_1^{\alpha_1}}\cdots\frac{\partial^{\alpha_m}}{\partial x_m^{\alpha_m}}=\frac{\partial^\alpha}{\partial x^\alpha}=D^\alpha.$$

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Thank you, Anon! Following you I have learned of falling factorial, the next time I will use them. – Giuseppe Apr 21 '12 at 18:20

Have you seen $\dfrac{\partial^2 y}{\partial x_1\,\partial x_2}$ before? This is the same thing as that but with more variables. It's called a "mixed partial derivative".

If, for example, $y$ is a function of four variables $x_1$, $x_2$, $x_3$, $x_4$, then we would have $$ \frac{\partial}{\partial x_2} \frac{\partial}{\partial x_4} \frac{\partial}{\partial x_4} \frac{\partial}{\partial x_1} y = \frac{\partial}{\partial x_2} \frac{\partial^2}{\partial x_4^2} \frac{\partial}{\partial x_1} y = \frac{\partial^{1+2+1} y }{\partial x_2\,\partial x_4^2\,\partial x_1} = \frac{\partial^4 y}{\partial x_2\,\partial x_4^2\,\partial x_1}.$$ In that case, $i_1=1$, $i_2=1$, $i_3=0$, and $i_4=2$, so the partial derivative above is $$ \frac{\partial^{i_1+i_2+i_3+i_4} y}{\partial x_1^{i_1}\,\partial x_2^{i_2}\,\partial x_3^{i_3}\,\partial x_4^{i_4}}. $$

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